Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $28.5$ years; the standard deviation is $5.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living between $23.2$ and $39.1$ years.
Solution: $28.5$ $23.2$ $33.8$ $17.9$ $39.1$ $12.6$ $44.4$ $95\%$ $68\%$ $13.5\%$ $13.5\%$ We know the lifespans are normally distributed with an average lifespan of $28.5$ years. We know the standard deviation is $5.3$ years, so one standard deviation below the mean is $23.2$ years and one standard deviation above the mean is $33.8$ years. Two standard deviations below the mean is $17.9$ years and two standard deviations above the mean is $39.1$ years. Three standard deviations below the mean is $12.6$ years and three standard deviations above the mean is $44.4$ years. We are interested in the probability of a zebra living between $23.2$ and $39.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the average lifespan. It also tells us that $68\%$ of the zebras will have lifespans within 1 standard deviation of the mean. The probability of a particular zebra living between $23.2$ and $39.1$ years is ${68\%} + \color{orange}{13.5\%}$, or $81.5\%$.